CIRCULAR MOTION AND GRAVITATIONAngular Measure, Angular Speed,

and Angular Velocity

BELLWORK 1. A tube is been placed upon the

1 m-high table and shaped into a three-quarters circle. A golf ball is pushed into the tube at one end at high speed. The ball rolls through the tube and exits at the opposite end. Describe the path of the golf ball as it exits the tube.

2. If the 50g golf ball leaves the tube with a velocity of 32 m/s at 45o, a) what is it’s maximum height, b) how long does it take to land, and c) what is the impulse force the ground exerts on the ball to bring it to a stop in 98 μs?

ALGEBRA 2 REVIEW An arc of a circle is a "portion" of the

circumference of the circle. The length of an arc is simply the length of its

"portion" of the circumference.  Actually, the circumference itself can be considered an arc length.

The length of an arc (or arc length) is traditionally symbolized by s.

The radian measure of a central angle of a circle is defined as the ratio of the length of the arc the angle subtends, s, divided by the radius of the circle, r.

ALGEBRA 2 REVIEW, CONT.

Relationship between Degrees and Radians:

When the arc length equals an entire circumference, we can use s = θr to get 2πr= θr and 2π = θ.

This implies that 2π = 360o

So and To change from degrees to radians,multiply by

To change from radians to degrees,multiply by

EXAMPLE PROBLEMS

1. Convert 50º to radians.2. Convert π/6 radians to degrees.3. How long is the arc subtended by an

angle of 7π/4 radians on a circle of radius 20.000 cm?

Answers: 1.5π/18 radians2.30 degrees3.109.96 cm

ANGULAR MEASURE

Circular motion is described using polar coordinates, (r, q)

x=rcosq and y = rsinq, where q is measured counterclockwise (ccw) from the positive x-axis.

Angle is defined as q = s/r, where s is the arc length, r is the radius, and q is the angle in radians. Also expressed as s=rq

Angular distance, Δ = - o is measured in degrees or radians. A radian is the angle that subtends an arc length that is equal to the radius (s=r)

1 rad = 57.3o , or 2rad = 360o

EXAMPLE 1

When you are watching the NASCAR Daytona 500, the 5.5 m long race car subtends an angle of 0.31o . What is the distance from the race car to you?

SOLUTION TO EXAMPLE 1

UNIFORM CIRCULAR MOTION IS THE MOTION OF AN OBJECT IN A CIRCLE WITH A CONSTANT OR UNIFORM SPEED.

When moving in a circle, an object traverses a distance around the perimeter of the circle

The distance of one complete cycle around the perimeter of a circle is known as the circumference

This relationship between the circumference of a circle, the time to complete one cycle around the circle, and the speed of the object is merely an extension of the average speed equation

COMBINING KINEMATICS AND CIRCULAR MEASUREMENT TO CALCULATE SPEED

The circumference of any circle can be computed using from the radius according to the equation

relating the speed of an object moving in uniform circular motion to the radius of the circle and the time to make one cycle around the circle (period, T), where R=radius:

Circumference = 2*pi*Radius

ANGULAR SPEED AND VELOCITY Instantaneous angular speed is the magnitude

of the instantaneous angular velocity

Tangential (linear) speed and angular speed are related to each other through

where r is the radius. The time it takes for an object to go through

one revolution is called the period, T. Then number of revolutions in one second is

called the frequency, f

THE DIRECTION OF THE VELOCITY VECTOR Velocity, being a

vector, has both a magnitude and a direction

Since an object is moving in a circle, its direction is continuously changing

direction of the velocity vector is tangential to the circular path

EXAMPLE 2

A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s. a) What is the angular speed of the

wheel?b) What is the tangential speed of a

point 0.10 m from the center of the wheel?

c) What is the period?d) What is the frequency?

ACCELERATION

an accelerating object is an object which is changing its velocity.

a change in either the magnitude or the direction constitutes a change in the velocity.

an object moving in a circle at constant speed is accelerating because the direction of the velocity vector is changing.

CHANGE IN VELOCITY The acceleration of the object is in the

same direction as the velocity change vector

Objects moving in circles at a constant speed accelerate towards the center of the circle.

CENTRIPETAL ACCELERATION

The linear tangential velocity vector changes direction as the object moves along the circle.

This acceleration is called centripetal acceleration (center-seeking) because it is always directed toward the center of the circle.

The magnitude of centripetal acceleration is given by

CENTRIPETAL FORCE

From Newton’s Second Law, we conclude that there MUST be a net force associated with centripetal acceleration.

Centripetal force is always directed toward the center of the circle since the net force on an object is in the same direction as acceleration.

EXAMPLE 3 A car of mass 1500 kg is negotiating a

flat circular curve of radius 50 m with a speed of 20 m/s.a) What is the source of the centripetal

force on the car? Explainb) What is the magnitude of the

centripetal acceleration of the car?c) What is the magnitude of the

centripetal force on the car?d) What is the minimum coefficient of

static friction between the car and the curve?

WARM-UP Convert the following angles from

degrees to radians or from radians to degrees, to two significant figures.-285o

195o

-90o

-4/3 π-270o

-3/4 π165o

π/3

CONVERT THE FOLLOWING ANGLES FROM DEGREES TO RADIANS OR FROM RADIANS TO DEGREES, TO TWO SIGNIFICANT FIGURES.

-285o

195o

-90o

-4/3 π-270o

-3/4 π165o

π/3

-5 rad or 1.6 π

3.4 rad or 1.08 π-1.6 rad or π/2

rad-240o or4/3 π rad -4.7 rad or 3/2 π

rad-135o or 3/4 π rad2.9 rad or 0.92 π

rad60o or π/3 rad

EXAMPLE 4: AFTER CLOSING A DEAL WITH A CLIENT, KENT LEANS BACK IN HIS SWIVEL CHAIR AND SPINS AROUND WITH A FREQUENCY OF 0.5 HZ. WHAT IS KENT’S PERIOD OF SPIN?

Given: f = 0.5 Hz Solve: T = 1/f T= 1/0.5 Hz T= 2 s

EXAMPLE 5: CURTIS’ FAVORITE DISCO RECORD HAS A SCRATCH 12 CM FROM THE CENTER THAT MAKES THE RECORD SKIP 45 TIMES EACH MINUTE. WHAT IS THE LINEAR SPEED OF THE SCRATCH AS IT TURNS?

The record makes 45 revolutions every minute (60 s), so T = 60 s/45 rev. = 1.3 s

r = 12 cm v = 2πr/T = 2π(12cm)/1.3s = 58 cm/s

EXAMPLE 6: MISSY’S FAVORITE RIDE AT THE TOPSFIELD FAIR IS THE ROTOR , WHICH HAS A RADIUS OF 4.0 M. THE RIDE TAKES 2.0 S TO MAKE ON E FULL REVOLUTION.A) WHAT IS MISSY’S LINEAR SPEED ON THE ROTOR?B) WHAT IS MISSY’S CENTRIPETAL ACCELERATION ON THE ROTOR? Given: r = 4.0 m, T = 2.0s v=2pir/T = 2pi(4.0m)/2.0s = 13 m/s ac = v2/r = 132/4.0 m = 42 m/s2

THE CAUSE OF CENTRIPETAL FORCE

In order to have an acceleration, there MUST be a force

What provides that force?

TensionApplied ForceFrictionSpring ForceGravitational Force

WHAT HAPPENS WHEN THE FORCE VANISHES?

MOTION IN A HORIZONTAL CIRCLE

The speed at which the object moves depends on the mass of the object and the tension in the cord

The centripetal force is supplied by the tension

CIRCULAR MOTION ABOUT A CONICAL PENDULUM

The object is in equilibrium in the vertical direction and undergoes uniform circular motion in the horizontal direction

v is independent of m

HORIZONTAL (FLAT) CURVE

The force of static friction supplies the centripetal force

The maximum speed at which the car can negotiate the curve is

BANKED CURVE

There is a component of the normal force that supplies the centripetal force

FICTIONAL FORCES

From the frame of the passenger (b),a force appears to push her toward the door

From the frame of the Earth, the car applies a leftward force on the passenger

The outward force is often called acentrifugal force

It is a fictitious force due to theacceleration associated with the car’s change in direction

THE GREAT MISCONCEPTION

Centrifugal, not to be confused with centripetal, means away from the center or outward.

Circular motion leaves the moving person with the sensation of being thrown OUTWARD from the center of the circle rather than INWARD

It’s really just inertia! http://www.physicsclassroom.com/mme

dia/circmot/cf.cfm

EFFECTS OF “PRETEND” FORCES

Although fictitious forces are not real forces, they can have real effects

Examples: Objects in the car do slide You feel pushed to the outside of a

rotating platform The Coriolis force is responsible for the

rotation ofweather systems and ocean currents

LOOP – THE – LOOP – A VERTICAL CIRCLE

At the bottom of the loop, the upward force experienced by the object is greater than it’s weight

Centripetal Force Vector ADDS to Normal Force Vector

LOOP – THE – LOOP

At the top of the circle, the force exerted on the object is less than its weight

Centripetal Force Vector TAKES AWAY from the normal force

WARM-UP:

Captain Chip, the pilot of a 60500 kg jet plane, is told he must remain in a holding pattern over the airport until it is his turn to land. If Captain Chip flies his plane in a circle whose radius is 50.0 km once every 30.0 min, what centripetal force must the air exert against the wings to keep the plane moving in a circle?

EXAMPLE 7.45

Many racetracks have banked turns, which allow the cars to travel faster around the curves than if the curves were flat. Actually, cars could also make turns on these banked curves if there were no friction at all.

Use a free body diagram to explain how this is possible.

EXAMPLE 7.46 An indy car with a speed of 120 km/hr

goes around a level, circular track with a radius of 1.00 km. What is the centripetal acceleration of the car?  

A. 1.11 m/s2

B. 0.555 m/s2

C. 3.49 m/s2

D.7.54 m/s2

.

EXAMPLE 7.52

A car with a constant speed of 83.0 km/hr enters a circular flat curve with a radius of curvature of 0.400 km. If the friction between the road and the car’s tires can supply a centripetal acceleration of 1.25 m/s2 does the car negotiate the curve safely? Justify your answer.

A. YesB. No

EXAMPLE 7.53 A student is to swing a bucket

of water in a vertical circle without spilling any (Fig. 7.29). (a) Use a free body diagram to help explain how this task is possible (what provides the centripetal force?). (b) If the distance from his shoulder to the center of mass of the bucket of water is 1.0 m, what is the minimum speed required to keep the water from coming out of the bucket at the top of the swing?

EXAMPLE 7.58 For a scene in a movie, a

stunt driver drives a 1.50 x 103 kg SUV with a length of 4.25 m around a circular curve with a radius of curvature of 0.333 km (Fig. 7.31). The vehicle is to be driven off the edge of a gully 10.0 m wide, and land on the other side 2.96 m below the initial side. What is the minimum centripetal acceleration the truck must have in going around the circular curve to clear the gully and land on the other side?  

BELLWORK: BONNIE IS ICE SKATING AT THE OLYMPIC GAMES. SHE IS MAKING A SHARP TURN WITH A RADIUS OF 22.6 M AND WITH A SPEED OF 16.1 M/S. USE NEWTON'S SECOND LAW TO DETERMINE THE ACCELERATION AND THE ANGLE OF LEAN OF BONNIE'S 55.0-KG BODY.

Given Info:m = 55.0 kgv = 16.1 m/sr = 22.6 mFind:a = ???Angle of lean = ???

SOLUTION TO BELLWORK

ac = v2/R a = (16.1 m/s)2/(22.6 m) = 11.5 m/s2

Fx = Fnet = m•a Fx

= (55.0 kg)•(11.5 m/s/s)

= 631 N

ANGULAR ACCELERATION: NON-UNIFORM CIRCULAR MOTION

Angular acceleration = the rate of change of angular velocity

EXAMPLE 8: A ROTATING CD

A CD accelerates uniformly from rest to its operational speed of 500 rpm in 3.50 s. A. What is the angular acceleration of

the CD during this time?B. What is the angular velocity of the

CD after this time? The angular acceleration after this time?

C. If the CD comes uniformly to a stop in 4.50 s, what is its angular acceleration during that part of the motion?

1. WHAT COULD THE POSITION AND VELOCITY VECTORS FOR THE LADY BUG LOOK LIKE?

A.

B.

C.

D.

2. WHAT COULD THE ACCELERATION AND VELOCITY VECTORS LOOK LIKE?

A.

B.

C.

D.

3. WHAT COULD THE POSITION & ACCELERATION VECTORS LOOK LIKE?

A.

B.

C.

D.

THE ACCELERATION WOULD NOT BE RADIAL OR THE PATH WOULD BE CIRCULAR. THIS IS VERY DIFFICULT TO SEE IN THE SIM.

X

Y

4. IF YOU HAD TWO BUGS MOVING IN CIRCLES LIKE THIS, WHAT COULD THE VELOCITY VECTORS AT POINT X VS POINT Y LOOK LIKE?

X YAB

C

D Any of the above

E None of the above are possible

IF THEY WERE CONNECTED WITH A BAR SO THEY HAD TO GO AROUND TOGETHER, IT WOULD BE LIKE IN LADYBUG REVOLUTION, BUT OTHERWISE THERE IS NO WAY TO KNOW THE LENGTH RELATIONSHIP, BUT THE VECTORS WOULD BE PARALLEL

X

Y

SEE FIG 7.14, P. 230

The acceleration and force have tangential components

Fc produces the centripetal acceleration

Fτ produces the tangential acceleration

ΣF = ΣFc + ΣFτ

TANGENTIAL ACCELERATION The tangential acceleration is the rate of

change of tangential velocity

For non-uniform circular motion, there are angular and tangential acceleration.

The total acceleration is the vector sum of the tangential and centripetal accelerations.

EXAMPLE 7.60 THE ANGULAR ACCELERATION IN CIRCULAR MOTION :

(a) is equal in magnitude to the tangential acceleration divided by the radius

(b) increases the angular velocity if both angular velocity and angular acceleration are in the same direction

(c) has units of s-2 (d) all of the preceding.  

EXAMPLE 7.62: FOR UNIFORM CIRCULAR MOTION,

(a) α = 0 (b) ω = 0 (c) r = 0 (d) none of the preceding.

EXAMPLE 7.66

During an acceleration, the angular speed of an engine increases from 600 rpm to 2500 rpm in 3.0 s. What is the average angular acceleration of the engine?  

TORQUE: MEASUREMENT OF THE TENDENCY OF A FORCE TO PRODUCE A ROTATION ABOUT AN AXIS.

Torque = perpendicular force x lever arm

τ=F x d d = distance from the pivot point, or

fulcrum, to the point where the component of the force perpendicular to the lever arm is being exerted

The longer the lever arm, the larger the torque

Counterclockwise rotation = POSITIVE τ Clockwise rotation = NEGATIVE τ Units = Newton x meter = Nm (but

NOT equal to Joules because torque is not a form of energy)

NOTE: weight is a force that can produce torque if the object is not supported at its center of gravity.

EXAMPLE 1

Ned tightens a bolt in his car engine by exerting 12 N of force on his wrench at a distance of 0.40 m from the fulcrum. How much torque must Ned produce to turn the bolt?

EXAMPLE 2: Mabel and Maude are seesawing on the

school playground and decide to see if they can move to the correct location to make the seesaw balance. Mabel weighs 400 N and she sits 2.00 m from the fulcrum of the seesaw. Where should 450-N Maude sit to balance the seesaw?

ANGULAR MOMENTUM: MEASURE OF HOW DIFFICULT IT IS TO STOP A ROTATING OBJECT

Moment of Inertia: The resistance of an object to changes in its rotational motion

moment of inertia = (mass)(radius)2

I = mr2

SI Units = kilogram meter squared = kgm2

An object that is rotating tends to continue to spinning at a constant rate unless acted on by an unbalanced force to alter that rotation

MOMENT OF INERTIA FOR VARIOUS SHAPES

Hoop rotating about its center

I=mr2

Hoop rotating about its diameter

I = ½ mr2

Solid cylinder I = ½ mr2

Stick rotating about its center of gravity

I = 1/12 ml2

Stick rotating about its end I = 1/3 ml2

Solid sphere rotating about its center of gravity

I = 2/5 mr2

ANGULAR MOMENTUM

Angular momentum = (mass)(velocity)(radius)

L = mvr SI Units = kilogram meter2 per second2

kgm2/s2

The product of the mass and the velocity for an object rotating at a distance, r, from the axis.

Angular Momentum of a system is conserved when no outside forces are acting

EXAMPLE 3

On the Wheel of Fortune game show, Ian spins the 15.0 kg wheel that has a radius of 1.40 m. What is the moment of inertia of this disk-shaped wheel? (thin cylinder)

EXAMPLE 4

Jun is twirling her 0.60 m majorette’s baton that has a mass of 0.40 kg. What is the moment of inertia of the baton as it spins about its center of gravity?

EXAMPLE 5At Wellesley College in Massachusetts there is a favorite tradition called hoop rolling. In their caps and gowns, seniors roll wooden hoops in a race in which the winner is said to be the first in the class to marry. Hilary rolls her 0.2 kg hoop across the finish line. The moment of inertia of the hoop is 0.032 kgm2 . What is the radius of the hoop?

EXAMPLE 6

Jupiter orbits the sun with a speed of 2079 m/s at an average distance of 71,398,000 m. If Jupiter has a mass of 1.90 x 1027 kg, what is its angular momentum as it orbits?